Question Number 63060 by rajesh4661kumar@gamil.com last updated on 28/Jun/19
Answered by Hope last updated on 28/Jun/19
$$\sqrt{\frac{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{2}} }{{sin}^{\mathrm{2}} \theta}}\: \\ $$$$=\mid\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}\mid \\ $$$$=\frac{\mid\mathrm{1}+{cos}\theta\mid}{\mid{sin}\theta\mid} \\ $$$${when}\:\:\:\pi>\theta>\mathrm{0}\:\:{so}\:{sin}\theta=+{ve} \\ $$$${cos}\theta<\mathrm{0}\:\:\:{but}\:\mid{cos}\theta\mid<\mathrm{1}\:\:\mid\mathrm{1}+{cos}\theta\mid=\mathrm{1}+{cos}\theta \\ $$$${so}\:\:\frac{\mid\mathrm{1}+{cos}\theta\mid}{\mid{sin}\theta\mid} \\ $$$$=\frac{\mathrm{1}+{cos}\theta}{{sin}\theta}={cosec}\theta+{cot}\theta \\ $$$$\boldsymbol{{but}}\:\boldsymbol{{when}}\:\:\mathrm{2}\pi>\theta>\pi \\ $$$$=\frac{\mid\mathrm{1}+{cos}\theta\mid}{\mid{sin}\theta\mid} \\ $$$$=\frac{\mathrm{1}+{cos}\theta}{−{sin}\theta} \\ $$$$=−{cosec}\theta−{cot}\theta \\ $$$$ \\ $$